Northfield Mount Hermon School
Advanced Placement Calculus
Problems in the Use of the Definite Integral
Dick Peller
Winter 1999
PROBLEM 1.
Oil is leaking from a tanker at the rate of R(t) = 2000 e-0.2t gallons per hour.
How much oil has leaked out of the tanker after 10 hours?
SOLUTION:
10
0
R(t )dt =
10
0
2000e −0.2t dt ≈ 8,647 gallons
PROBLEM 2.
The EPA was recently asked to investigate a spill of radioactive iodine. Measurements
showed the ambient radiation levels at the site to be four times the maximum acceptable
level, so the EPA ordered an evacuation of the surrounding neighborhood.
It is known that the level of radiation from an iodine source decreases according to the
formula R(t) = R0 e-0.004t, where R(t) is the radiation level (in millirems/hour) a time t,
and R0 is the initial radiation level, and t is time in hours.
a) How long will it take the site to reach an acceptable level of radiation?
b) How much total radiation (in millirems) will have been emitted by this time,
assuming the maximum acceptable limit is 0.6 millirems/hour?
SOLUTION:
a) e-0.004t = .25, so t = 347 hours.
b)
2.4e − 0.004t dt = 450 millirems
Page 1 of 9
PROBLEM 3.
Suppose the density of a circular oil slick on the surface of the water is given by
100
ρ (r ) =
kg / m2 .
2
1+ r
a) Given that the slick extends from r=0 to r=1000 meters, write a Riemann sum that
approximates the total mass of oil.
b) Determine the total mass of the oil.
c) Within what radius is 75% of the mass?
SOLUTION:
( Area )( Density ) ≈
a) Mass =
b)
200πr
dr ≈ 4,340 kg
1+ r2
Mass =
c)
k
0
100
(2πri ) ∆ri
2
i =1 1 + ri
200πr
dr ≈ (0.75)(4340) kg , k ≈ 178 meters
1+ r2
PROBLEM 4.
Suppose the density of cars, in cars/mile, along a 30 mile stretch of the Mass Pike (during
rush hour) can be modeled by ρ ( x ) = 100(2 − 3 01
. x + 0.2 ) , where x represents the number
of miles from Boston.
a) Write a function that gives the number of cars from Boston to a point x miles from
Boston.
b) Use this function to determine the total number of cars on this 30-mile stretch of road.
SOLUTION:
a) C(x) =
b) C(30) =
x
0
ρ (t )dt
30
0
100(2 − 3 0.1x + 0.2 )dt ≈ 2,551 cars
Page 2 of 9
PROBLEM 5.
Greater Boston can be approximated by a semi-circle of radius 8 miles from its center on
the coast. Moving away from the center along a radius, the population density starts to
decrease according to the data given in the table, where ρ (r ) , measured in thousands of
people/mile2, is the population density at a distance r miles from the center.
r (miles)
ρ (r )
0
75
1
75
2
67.5
3
60
4
52.5
5
45
6
37.5
7
30
8
22.5
Using this data and Riemann sums, estimate the total population living in the 8 mile
radius.
SOLUTION:
P(r) =
8
Area x Density ≈
i =0
or
Area x Density ≈
9
i =1
πri ρ (ri ) ∆ri = 3,958 thousand people
πri ρ (ri ) ∆ri
= 4,524
thousand people.
PROBLEM 6.
A faucet is turned on at time t = 0, and t minutes later water is flowing into a barrel at a
rate of R(t) = t2 + 4t gallons/minute, for t in [0,5].
a)
b)
c)
d)
How much water was added to the barrel during these five minutes?
How much water was added to the barrel during the third minute of the flow?
Find the average rate of flow for the first five minutes.
Find a one-minute interval in which the total flow was equal to the average flow.
SOLUTION:
a) V(5) – V(0) =
5
0
dV
dt =
dt
5
0
(t 2 + 4t )dt =
275
gallons
3
(t 2 + 4t )dt =
275 gallons
15 minute
b) V(3) – V(2) = 49/3 gallons.
c) Average rate of flow =
d) solve
x +1
x
(t 2 + 4t )dt =
1
5
5
0
275
We find x = 2.22, so the interval is [2.22,3.22]
15
Page 3 of 9
PROBLEM 7.
A mold of varying thickness is growing in the shape of a disk; the mold is thickest
at the center of the disk and the thickness decreases as the distance from the
center of the mold increases. At a distance R millimeters from the center, the
thickness of the mold is 4- 0.3eR millimeters, for 0 ≤ R ≤ 2.
In the figure above, circles have been drawn of radii 0.5, 1.0, 1.5 and 2.0
millimeters. The volume of the mold in the middle ring can be approximated in
the following way: First we approximate the area of the ring by the product of the
outer circumference and the thickness of the ring.
. )(0.5) .
Thus, Area of ring ≈ 2πR ∆R = 2π (15
Next we approximate the volume by using the height of the mold on the outer
circumference. Note that this height is constant on such a circle.
height = 4 – 0.3e1.5.
Finally, the volume of the mold in the middle ring is given by the height times the
area, so Volume = (4 – 0.3e1.5) 2π (1.5)(0.5).
Using this technique with the other three subregions in the given figure,
approximate the volume of the mold.
SOLUTION:
Volume =
(4 – 0.3e0.5) 2π (0.5)(0.5) + (4 – 0.3e1.0) 2π (1.0)(0.5) +
(4 – 0.3e1.5) 2π (1.5)(0.5) + (4 – 0.3e2.0) 2π (2.0)(0.5)
c) The answer to part a) is a Riemann sum with n = 4 over the interval [0,2] for
what definite integral?
SOLUTION:
2
0
(4 − 0.3e R )2πR dR
Page 4 of 9
d) Use this integral to find the volume of the solid.
SOLUTION:
Volume = 34.45 cubic millimeters
d) Use the same technique to set up a definite integral that would give the total number of
people, in thousands, living in Diskville, if the population R miles from the center of
12
Diskville is
thousand people per square mile, and Diskville has a circular
2+R
boundary 5 miles from the center of the town.
SOLUTION:
5
0
12
(2πR ) dR ≈ 188,079 people
2+ R
PROBLEM 8.
The value V of a Tiffany lamp, worth $225 in 1965, increases at the rate of 15% per year.
Its value t years after 1965 is given by V(t) = 225(1.15)t.
Find the average value of the lamp over the period 1965–2000.
SOLUTION:
average value =
1
35
35
0
225(1.15) t dt
Page 5 of 9
PROBLEM 9.
A bar of metal is cooling from 1000o C to room temperature, which is 20oC.
The temperature, H, of the bar t hours after it starts cooling is given by
H(t) = 20 + 980 e-0.1t.
a) Find the temperature of the bar after one hour.
b) Find the average value of the temperature of the bar over the first hour.
c) Is your answer to part b) greater or smaller than the average of the temperatures at the
beginning and end of the hour? Explain your answer in terms of the concavity of the
graph of H(t).
SOLUTION:
a) H(t) = 20 + 980 e-0.1(1) = 906.74
b) average value of H over [0,1] =
c)
1 1
(20 + 980e − 0.1t )dt ≈ 952.59°C
0
1− 0
H (0) + H (1) 1000 + 906.74
=
= 953.37 . The function is concave up.
2
2
PROBLEM 10.
After t hours, a population of bacteria is growing at the rate of 2t million bacteria per
hour. Estimate the total increase in the bacteria population during the first hour.
SOLUTION:
Change in population =
1
0
2 t dt ≈ 1.4 million bacteria .
PROBLEM 11.
The air density in kg/m3 h meters above the surface of the earth is given by P = f(h).
Find the mass of a cylindrical column of air 2 meters in diameter and 25 km high.
SOLUTION:
The mass of a horizontal air slice with thickness ∆h ≈ (Volume)( Density ) = (π ∆h) f (hi ) .
So the total mass ≈
definite integral
n −1
π f (hi ) ∆h kg. As n approaches infinity, this sum represents the
i=0
25000
0
π f (h) dh kg.
Page 6 of 9
PROBLEM 12.
Water leaks out of a tank through a square hole with 1 inch sides. At time t (in seconds),
the velocity of water flowing through the hole is v = g(t) ft/sec.
What is the definite integral that represents the total amount of water lost in the first
minute?
SOLUTION:
The area of the square hole is 1/144 square feet. The amount of water which has passed
through in some time ∆t can be thought of as a rectangular solid whose volume V =
(Area)(Height) = (Area)(Velocity)(Time).
Over ∆t seconds, the amount of water lost
1
1
≈
ft 2 ( g (t ) ft / sec) (∆t sec) =
g (t ) ∆t ft 3 .
144
144
Therefore, the total amount of water lost ≈
sum goes to the definite integral
60
0
1
g (t ) dt ft 3 , and as ∆t goes to 0, this
144
1
g (t ) dt .
144
PROBLEM 13.
The rate of consumption of oil in the United States since 1974 can be modeled by the
function C(t) = 21.3 e0.04t, where t is the number of years since 1974 and C(t) is measured
in billions of barrels of oil per year.
Based on this model, how much oil was used in the United States in the decade beginning
in 1980 and ending 1989?
SOLUTION:
Amount of oil = lim
∆t →∞
15
t =6
C ( t ) ∆t =
15
6
21.3 e 0.04t dt ≈ 332.9 billion gallons of oil
Page 7 of 9
PROBLEM 14.
When a body moves a distance d along a straight line as the result of being acted upon by
a force that has constant magnitude F in the direction of the motion, the work done by the
force in moving the body is F times d: W = Fd.
A leaky 5 pound bucket is lifted from the ground into the air by a worker pulling in 20
feet of rope at a constant speed. The rope weighs 0.008 lb/ft. The bucket starts with 2
gallons of water (16 lbs.) and leaks at a constant rate. It finishes draining just as it reaches
the top.
How much work is done in:
a) lifting the water alone?
b) lifting the water and the bucket?
c) lifting the water, the bucket and the rope?
SOLUTION:
a) When the bucket is x ft. off the ground, the water weighs
20 − x
4x
F ( x) = 16
= 16 −
lb.
20
5
The work done is W =
20
0
F ( x)dx =
20
0
16 −
4x
dx = 160 ft lb.
5
b) To lift the bucket and the water, the work is 160 + (5)(20)=260 ft. lb.
c) The work in lifting all three:
F(x) = 16 −
4x
+ 5 + (0.08)(20 − x) , so the total work is 276 ft. lb.
5
Page 8 of 9
PROBLEM 15.
How much work does it take to pump the water from and upright and full right circular
cylindrical tank of radius 5 ft. and height 10 ft. to a level 4 ft. above the top of the tank?
SOLUTION:
Consider moving "thin" slabs of water, with thickness ∆y.
The volume of a slab is about ∆V = π (radius)2 (thickness) = π (52) (∆y) = 25π ∆y ft3.
The force required to lift this slab is its weight,
F(y) = w ∆V = 25πw(∆y) lb.,
where w is the weight of one cubic foot of water.
The distance through which F(y) must act is about (14-y) feet, so the work done in lifting
this slab 4 feet above the top of the tank is about
∆W = 25 π w (14-y) ft. lb.
So the total work is given by the definite integral
Work =
10
0
25πw(14 − y )dy = 2,250 πw ft.lb.
Page 9 of 9